256 如何存储在 char 变量和 unsigned char 中

how 256 stored in char variable and unsigned char(256 如何存储在 char 变量和 unsigned char 中)

问题描述

最多 255，我可以理解整数是如何存储在 char 和 unsigned char 中的；

Up to 255, I can understand how the integers are stored in char and unsigned char ;

#include<stdio.h>
int main()
{
        unsigned char a = 256;
        printf("%d
",a);
        return(0);
}

在上面的代码中，我为 unsigned char 和 char 输出了 0.

In the code above I have an output of 0 for unsigned char as well as char.

对于256我认为是代码中整数的存储方式(这只是猜测):

For 256 I think this is the way the integer stored in the code (this is just a guess):

第一个 256 转换为二进制表示，即 100000000(共 9 位).

First 256 converted to binary representation which is 100000000 (totally 9 bits).

然后他们删除删除 最左边的位(设置的位)，因为 char 数据类型只有 8 位内存.

Then they remove the remove the leftmost bit (the bit which is set) because the char datatype only have 8 bits of memory.

所以它在内存中存储为 00000000 ，这就是为什么它打印 0 作为输出.

So its storing in the memory as 00000000 , that's why its printing 0 as output.

猜测正确还是有其他解释?

Is the guess correct or any other explanation is there?

推荐答案

你猜对了.转换为无符号类型使用模运算:如果值超出范围(太大或负数)，则以 2^N 为模减少，其中 N 是目标类型中的位数.因此，如果(通常情况下)char 有 8 位，则该值以 256 为模减少，因此 256 变为 0.

Your guess is correct. Conversion to an unsigned type uses modular arithmetic: if the value is out of range (either too large, or negative) then it is reduced modulo 2^N, where N is the number of bits in the target type. So, if (as is often the case) char has 8 bits, the value is reduced modulo 256, so that 256 becomes zero.

请注意，转换为有符号类型没有这样的规则 - 超出范围的值会给出实现定义的结果.另请注意，char 未指定为正好 8 位，在不太主流的平台上可以更大.

Note that there is no such rule for conversion to a signed type - out-of-range values give implementation-defined results. Also note that char is not specified to have exactly 8 bits, and can be larger on less mainstream platforms.

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